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3 years ago

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A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

n of gravity is 9.8 m/s 2 . Answer in units of N.

1 answer:

pickupchik [31]3 years ago

4 0

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

$m=\dfrac{W}{g}$

$m=\dfrac{110\ N}{9.8\ m/s^2}$

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

$M=d\times V$

d is the density of water, $d=1000\ kg/m^3$

V is the volume of rock, $V=0.00337\ m^3$

$M=1000\ kg/m^3\times 0.00337\ m^3$

M = 3.37 kg

The apparent weight in water, W = m - M

$W=7.85\ kg\times 9.8\ m/s^2$

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

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